A magazine wants to know the proportion of the public that would favour a candidate for the Prime Minister position. If a random sample of 1,600 is chosen with a confidence level of 95%, what is the largest possible margin of error?
Question
A magazine wants to know the proportion of the public that would favour a candidate for the Prime Minister position. If a random sample of 1,600 is chosen with a confidence level of 95%, what is the largest possible margin of error?
Solution
To calculate the margin of error for a proportion in a population, you can use the formula:
Margin of Error = Z * sqrt[(p(1-p))/n]
Where:
- Z is the z-score, which corresponds to the desired confidence level. For a confidence level of 95%, the z-score is approximately 1.96.
- p is the estimated proportion of the population which has the characteristic of interest. Since we don't have this information, we'll use the value that maximizes the product p(1-p), which is 0.5.
- n is the size of the sample, which is 1,600 in this case.
Substituting these values into the formula, we get:
Margin of Error = 1.96 * sqrt[(0.5(1-0.5))/1600] = 1.96 * sqrt[(0.25)/1600] = 1.96 * sqrt[0.00015625] = 1.96 * 0.0125 = 0.0245 or 2.45%
So, the largest possible margin of error for this survey is 2.45%.
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